What is a equivalent expression for 5w-15

Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)

ehidna [41]

This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

a)

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ], slope is 86.3

111) for [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope = ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11) [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111) [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

b)

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

3 0

3 years ago

Whats the action “verb of the boy runs into the woods”? Please help

Brilliant_brown [7]

Answer:

the boy runs onto the woods

Step-by-step explanation:

verb: a word that describes an action

so the answer is runs

4 0

3 years ago

I will brainlest this don't answer if out the real answe thank you

Mashutka [201]

Answer:

1) Product of 2 x 3/5 is 6/5 or 1 and 1/5.

2) 20/9 or 2 and 2/9

3) 15/8 or 1 and 7/8

4) 15/6 or 2 and 3/6 or 2 and 1/2

5) 6/4 or 1 and 2/4 or 1 and 1/2

Step-by-step explanation:

You multiply the whole number by the numerator (top number) and then divide the result by the denominator(bottom number). Afterward, simplify the fraction if you can, for example, 6/9. Both the numerator and denominator can be divided by 3. Therefore, the answer to that example is 2/3.

7 0

3 years ago

Read 2 more answers

A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their fou

Jet001 [13]

Answer:

We accept H₀. We don´t have enough evidence to support the difference between the two means

Step-by-step explanation:

Group 1 : Comunity College

Sample:

size : n₁ = 54

Sample mean x₁ = 947

Sample standard deviation s₁ = 254

Group 2 : Local University

Sample

size : n₂ = 66

Sample mean : x₂ = 1011

Sample standard deviation s₂ = 87

Significance level α = 0,05 α = 5 % CI = 95 %

Both samples are with n> 30 so we can use z table for the test

from z-table we find z(c) = 1,96

The problem statement establishes to run a hypothesis test to determine if the means are statistically the same ( or different) then we use two-tail-test

Hypothesis Test

Null Hypothesis H₀ x₁ - x₂ = 0 or x₁ = x₂

Alternative hypothesis Hₐ x₁ - x₂ ≠ 0 or x₁ ≠ x₂

To calculate

z(s) = ( x₁ - x₂ ) / √ s₁²/n₁ + s₂²/n₂ )

z(s) = 947 - 1011 / √ (254)²/54 + (87)² / 66

z(s) = - 64 /√ 1194,74 + 114,68

z(s) = - 64 / 36,185

z(s) = - 1,77

Comparing modules of |z(c)| and |z(s)|

z(s) < z(c)

Then z(s) is in the acceptance region we accept H₀ . There is not enough evidence to support differences in the means of the two groups

3 0

3 years ago

What is the solution to the system of linear equations?<br><br> −9x+4y=55<br><br> −11x+7y=82

LuckyWell [14K]

Here is my process with mathematical expression interpreter, LaTeX. Full process given below to obtain the values of variable "x" and "y" altogether, solutions to these system of linear equations.

$\begin{bmatrix}-9x & + & 4y & = & 55 & \bf{- - - \: Eq. \: 1} \\ \\ -11x & + & 7y & = & 82 & \bf{- - - \: Eq. \: 2} \end{bmatrix}$

Now here, we should isolate a variable, or take it as a separate form to find the equation, and furthermore substitute the value of variable "y" into the original isolation of "x", to obtain both the solutions for this linear system of equation. Perform this on equation number 1 (Eq. 1).

Subtract the variable attached value by "4y" on both the sides, in current expression.

$\mathbf{-9x + 4y - 4y = 55 - 4y}$

$\mathbf{-9x = 55 - 4y}$

Both the sides, perform a division of value "-9".

$\mathbf{\dfrac{-9x}{-9} = \dfrac{55}{-9} - \dfrac{4y}{-9}}$

$\mathbf{x = \dfrac{55 - 4y}{-9}}$

$\mathbf{x = - \dfrac{55 - 4y}{9}}$

Substitute or just plug the value of newly obtained expression for variable "x" into Equation, numbered as "2" (Eq. 2.) and isolate further for the variable "y", to obtain first solution for this linear equation.

$\mathbf{-11 \Bigg(- \dfrac{55 - 4y}{9} \Bigg) + 7y = 82}$

$\mathbf{\dfrac{(55 - 4y) \times 11}{9} + 7y = 82}$

Multiply both the sides by a value of "9".

$\mathbf{\dfrac{(55 - 4y) \times 11}{9} + 7y \times 9 = 82 \times 9}$

$\mathbf{11 (55 - 4y) + 63y = 738}$

$\mathbf{605 - 44y + 63y = 738}$

$\mathbf{605 + 19y = 738}$

Subtract both the sides by a value of "- 605".

$\mathbf{605 + 19y - 605 = 738 - 605}$

$\mathbf{19y = 133}$

Divide both the sides by "19".

$\mathbf{\dfrac{19y}{19} = \dfrac{133}{19}}$

$\boxed{\mathbf{y = 7}}$

Substitute this variable value of "y = 7" , into our original isolation for variable "x", the expression is to be substituted by that value to complete the solutions for the linear equations. That is:

$\mathbf{x = - \dfrac{55 - 4y}{9}; \quad y = 2}$

$\mathbf{\therefore \quad x = - \dfrac{55 - 4 \times 7}{9}}$

$\mathbf{\therefore \quad x = - \dfrac{55 - 28}{9}}$

$\mathbf{x = - \dfrac{27}{9}}$

$\boxed{\mathbf{x = - 3}}$

Finalised solutions for these linear system of equations for two components , is:

$\boxed{\mathbf{\underline{\therefore \quad Final \: Solutions \: for \: these \: System \: of \: Linear \: Equations: \: x = - 3, \: \: y = 7}}}$

Hope it helps.

5 0

3 years ago